[LeetCode] 700. Search in a Binary Search Treer_Easy_tag: Binary Search Tree

Given the root node of a binary search tree (BST) and a value. You need to find the node in the BST that the node's value equals the given value. Return the subtree rooted with that node. If such node doesn't exist, you should return NULL.

For example, 

Given the tree:
        4
       / \
      2   7
     / \
    1   3

And the value to search: 2

You should return this subtree:

      2     
     / \   
    1   3

In the example above, if we want to search the value 5, since there is no node with value 5, we should return NULL.

Note that an empty tree is represented by NULL, therefore you would see the expected output (serialized tree format) as [], not null.

 

这个是Binary Search Tree 中的一个basic skill，就是去找到tree中是否有某个点。因为binary search tree的特点，left child < node < right child, 所以可以判断是在left tree or right tree， 所以T：O（h）， h = lg2 (n).

T: O(h)    S: O(1)

Code

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def searchBST(self, root, val):
        """
        :type root: TreeNode
        :type val: int
        :rtype: TreeNode
        """
        #ans = None
        while root:
            if root.val > val:
                root = root.left
            elif root.val == val:
                return root
            else:
                root = root.right
        
        

 2) recursive

class Solution(object):
    def searchBST(self, root, val):
        """
        :type root: TreeNode
        :type val: int
        :rtype: TreeNode
        """
        if not root: return 
        if root.val == val:
            return root
        elif root.val < val:
            return self.searchBST(root.right, val)
        else:
            return self.searchBST(root.left, val)